Difference between revisions of "2014 AMC 10B Problems/Problem 2"

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We can factor the numerator and the denominator. The numerator becomes 2^3(1+1) and the denominator becomes 2^-3(1+1). The (1+1)'s cancel so we are left with 2^3 over 2^-3. This leaves us with 2^6 which equals to E) 64

Revision as of 23:48, 9 April 2019

Problem

What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$?

$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$

Solution

We can synchronously multiply ${2^3}$ to the polynomials both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{{64 (\textbf{E})}}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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We can factor the numerator and the denominator. The numerator becomes 2^3(1+1) and the denominator becomes 2^-3(1+1). The (1+1)'s cancel so we are left with 2^3 over 2^-3. This leaves us with 2^6 which equals to E) 64